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TERM PAPER

Engineering Mathematics-II

MTH-102

Topic: Application of exact differential equation in engineering .Using

equations of your own choice.

Submitted to: Submitted by:

Ms, Manreet singh Mr. Shailesh singh

Deptt. Of MATHEMATICS Roll. No.Rd4901A59

Reg.No10908518

ClassB.tech (ME)


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ACKNOWLEDGEMENT

I want to thanks God and my mother for their blessings. I am also

very grateful and thankful to Mr. Manreet singh (subject teacher)

and other authorities of the university to provide me such an

opportunity and guidance for completing my assigned term paper

work.

Last but not the least I want to thank my friends without whose

support my work would not be as easier as it was with their valuable

support.


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INDEX:-

01 Overview

02 Exact differential equation

03Method of solving of exact differential equatons

04 Application of exact differential equation

05 References

Overview:-

For one dimension, a differential

is always exact.

For two dimensions, in order that a differential

be an exact differential in a simply-connected region R of the xy-plane, it is necessary andsufficient that betweenA andB there exists the relation:

For three dimensions, a differential

is an exact differential in a simply-connected regionR of thexyz-coordinate system if between the

functionsA,B and Cthere exist the relations:

; ;


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These conditions are equivalent to the following one: If G is the graph of this vector valued

function then for all tangent vectors X,Y of thesurface G then s(X,Y)=0 with s the symplectic

form.

These conditions, which are easy to generalize, arise from the independence of the order ofdifferentiations in the calculation of the second derivatives. So, in order for a differential dQ, that

is a function of four variables to be an exact differential, there are six conditions to satisfy.

In summary, when a differential dQ is exact:

the function Q exists;

, independent of the path followed.

In thermodynamics, when dQ is exact, the function Q is a state function of the system. The

thermodynamic functions U, S,H,A and G are state functions. Generally, neitherworknorheatisa state function. An exact differential is sometimes also called a 'total differential', or a 'full

differential', or, in the study of differential geometry, it is termed an exact form.


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Exact Differential

A differential of the form

Is exact (also called a total differential) if is path-independent. This will be true if

so and must be of the form

But


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so

There is a special notation encountered especially often in statistical thermodynamics. Consider

an exact differential

Then the notation , sometimes referred to as constrained variable notation, means "the

partial derivative of with respect to with held constant." Extending this notation a bit leads to

the identity

Where it is understood that on the last-hand side is treated as a variable that can itself

be help constant.

METHODS FOR SOLVING EXACT DIFFERENTIAL EQUATIONS:-

STEP 1. Integrate M w.r.t x keeping y constant

STEP 2. Integrate w.r.t y, only those terms of N

Which do not contain x

STEP 3. Result of 1 + result of 2 = constant


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Mdx+ Ndy =0

M /y = N/x { exact differential equation}

.: the solution of Mdx+Ndy = 0 is

Mdx + (terms not containing x)dy = c

Y=c

THERE IS ANOTHER METHOD TO SOLVE THE EXACT

DIFFERENCIAL EQUATION:

Consider the equation

f(x,y) = C

Taking the gradient we get

fx(x,y)i + fy(x,y)j = 0

We can write this equation in differential form as

fx(x,y)dx+ fy(x,y)dy = 0

Now divide by dx (we are not pretending to be rigorous here) to get

fx(x,y)+ fy(x,y) dy/dx = 0

Which is a first order differential equation? The goal of this section is to go backward. That is if

a differential equation if of the form above, we seek the original function f(x,y) (called

apotentialfunction). A differential equation with a potential function is called exact. If you have

had vector calculus, this is the same as finding the potential functions and using the fundamental

theorem of line integrals.


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Example:

Solve

4xy + 1 + (2x2 + cos y) y = 0

Solution

We seek a function f(x,y) with

fx(x,y) = 4xy + 1 and fy(x,y) = 2x2

+ cos y

Integrate the first equation with respect to x to get

f(x,y) = 2x2y + x + C(y) Notice since y is treated as a constant,. we write C(y).

Now take the partial derivative with respect to y to get

fy(x,y) = 2x2 + C'(y)

We have two formulae for fy(x,y) so we can set them equal to each other.

2x2 + cos y = 2x2 + C'(y)

That is

C'(y) = cos y

or

C(y) = sin y

Hence


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f(x,y) = 2x2y + x + sin y

The solution to the differential equation is

2x2y + x + sin y = C

Applications of Exact Differential Equations


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1. Orthogonal trajectories :- The term here orthogonal meansperpendicular,

and trajectory meanspath orcurve. Orthogonal trajectories are two families of curves

that always intersect perpendicularly. A pair of intersecting curves will be perpendicular

if the product of their slopes is 1, that is, if the slope of one is the negative reciprocal

of the slope of the other. Since the slope of a curve is given by the derivative, two

families of curves 1(x,y, c) = 0 and 2(x,y, c) = 0 (where c is a parameter) will be

orthogonal wherever they intersect if

Example 1: The electrostatic field created by a positive point charge is pictured as a

collection of straight lines which radiate away from the charge (Figure 1 ). Using the fact

that the equipotential(surfaces of constant electric potential) are orthogonal

the electric field lines, determine the geometry of the equipotenitials of a point charge.

Figure 1

If the origin of axy coordinate system is placed at the charge, then the electric field lines

can be described by the family
http://www.cliffsnotes.com/study_guide/Applications-of-FirstOrder-Equations.topicArticleId-19736,articleId-19734.html#LeDuc3209c06-fig-0001http://www.cliffsnotes.com/study_guide/Applications-of-FirstOrder-Equations.topicArticleId-19736,articleId-19734.html#LeDuc3209c06-fig-0001


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The first step in determining the orthogonal trajectories is to obtain an expression for the

slope of the curves in this family that does not involve the parameterc. In the present case,

The differential equation describing the orthogonal trajectories is therefore

since the right-hand side of (**) is the negative reciprocal of the right-hand side of (*).

Because this equation is separable, the solution can proceed as follows:

where c2 = 2 c.

The equipotential lines (that is, the intersection of the equipotential surfaces with any plane

containing the charge) are therefore the family of circles x2 +y2 = c2 centered at the origin.

The equipotential and electric field lines for a point charge are shown in Figure 2.


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Figure 2

Example 2: Determine the orthogonal trajectories of the family of circlesx2 +

(y c)2 = c2 tangent to thex axis at the origin.

The first step is to determine an expression for the slope of the curves in this family that

does not involve the parameterc. By implicit differentiation,

To eliminatec, note that


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The expression fordy/dx may now be written in the form

Therefore, the differential equation describing the orthogonal trajectories is

since the right-hand side of (**) is the negative reciprocal of the right-hand side of (*).

If equation (**) is written in the form

note that it is not exact (since My = 2y but Nx = 2 y). However, because

is a function ofx alone, the differential equation has

as an integrating factor. After multiplying through by = x2, the differential equation

describing the desired family of orthogonal trajectories becomes


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Which is now exact (because My= 2x2y =Nx). Since

and

the solution of the differential equation is

(The reason the constant was written as 2 c rather than as c will be apparent in the

following calculation.) With a little algebra, the equation for this family may be rewritten:

This shows that the orthogonal trajectories of the circles tangent to the x axis at the origin

are the circles tangent to the y axis at the origin! See Figure 3 .


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Figure 3

2. Radioactive decay:-. Some nuclei are energetically unstable and can spontaneously

transform into more stable forms by various processes known collectively as radioactive

decay. The rate at which a particular radioactive sample will decay depends on the

identity of the sample. Tables have been compiled which list the half-lives of various

radioisotopes. The half-life is the amount of time required for one-half the nuclei in a

sample of the isotope to decay; therefore, the shorter the half-life, the more rapid the

decay rate.

The rate at which a sample decays is proportional to the amount of the sample present.

Therefore, ifx (t) denotes the amount of a radioactive substance present at time t, then

(The rate dx/ dtis negative, sincex is decreasing.) The positive constant kis called the rate

constant for the particular radioisotope. The solution of this separable first-order equation is


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wherexodenotes the amount of substance present at time t= 0. The graph of this equation

(Figure 4 ) is known as the exponential decay curve:

Figure 4

The relationship between the half-life (denoted T1/2) and the rate constant kcan easily be

found. Since, by definition,x = x6 at t= T1/2, (*) becomes

Because the half-life and rate constant are inversely proportional, the shorter the half-life,

the greater the rate constant, and, consequently, the more rapid the decay.

Radiocarbon dating is a process used by anthropologists and archaeologists to estimate the

age of organic matter (such as wood or bone). The vast majority of carbon on earth is

nonradioactive carbon-12 (12C). However, cosmic rays cause the formation of carbon-

14 (14C), a radioactive isotope of carbon which becomes incorporated into living plants (and


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therefore into animals) through the intake of radioactive carbon dioxide ( 14CO2). When the

plant or animal dies, it ceases its intake of carbon-14, and the amount present at the time of

death begins to decrease (since the14C decays and is not replenished). Since the half-life

of14C is known to be 5730 years, by measuring the concentration of14C in a sample, its age

can be determined.

Example 3: A fragment of bone is discovered to contain 20% of the usual 14C concentration.

Estimate the age of the bone.

The relative amount of14C in the bone has decreased to 20% of its original value (that is, the

value when the animal was alive). Thus, the problem is to calculate the value oftat whichx

(t) = 0.20xo (wherex = the amount of14C present). Since

The exponential decay equation (*) says

Newton's Law of Cooling. When a hot object is placed in a cool room, the object dissipates

heat to the surroundings, and its temperature decreases. Newton's Law of Cooling states that

the rate at which the object's temperature decreases is proportional to the difference between

the temperature of the object and the ambient temperature. At the beginning of the cooling

process, the difference between these temperatures is greatest, so this is when the rate of

temperature decrease is greatest. However, as the object cools, the temperature difference

gets smaller, and the cooling rate decreases; thus, the object cools more and more slowly as


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time passes. To formulate this process mathematically, let T (t) denote the temperature of

the object at time tand let Ts denote the (essentially constant) temperature of the

surroundings. Newton's Law of Cooling then says

Since Ts < T(that is, since the room is cooler than the object), Tdecreases, so the rate of

change of its temperature, dT/dt, is necessarily negative. The solution of this separable

differential equation proceeds as follows:

Example 4: A cup of coffee (temperature = 190F) is placed in a room whose temperature is

70F. After five minutes, the temperature of the coffee has dropped to 160F. How many

more minutes must elapse before the temperature of the coffee is 130F?

Assuming that the coffee obeys Newton's Law of Cooling, its temperature Tas a function of

time is given by equation (*) with Ts= 70:

Because T(0) = 190, the value of the constant of integration ( c) can be evaluated:


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Furthermore, since information about the cooling rate is provided (T = 160 at time t= 5

minutes), the cooling constant kcan be determined:

Therefore, the temperature of the coffee tminutes after it is placed in the room is

Now, setting T= 130 and solving fortyields


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This is the totalamount of time after the coffee is initially placed in the room for its

temperature to drop to 130F. Therefore, after waiting five minutes for the coffee to cool

from 190F to 160F, it is necessary to then wait an additional seven minutes for it to cool

down to 130F.

Skydiving. Once a sky diver jumps from an airplane, there are two forces that determine her

motion: the pull of the earth's gravity and the opposing force of air resistance. At high

speeds, the strength of the air resistance force (the drag force) can be expressed as kv2,

where v is the speed with which the sky diver descends and kis a proportionality constant

determined by such factors as the diver's cross-sectional area and the viscosity of the air.

Once the parachute opens, the descent speed decreases greatly, and the strength of the air

resistance force is given byKv.

Newton's Second Law states that if a net forceFnet acts on an object of mass m, the objectwill experience an acceleration a given by the simple equation

Since the acceleration is the time derivative of the velocity, this law can be expressed in the

form

In the case of a sky diver initially falling without a parachute, the drag force isFdrag = kv2,

and the equation of motion (*) becomes

Or more simply,


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Where b = k/m. [The lettergdenotes the value of the gravitational acceleration, and mgis

the force due to gravity acting on the mass m (that is, mgis its weight). Near the surface of

the earth,gis approximately 9.8 meters per second2.] Once the sky diver's descent speed

reaches v = , the preceding equation says dv/ dt= 0; that is, v stays constant.

This occurs when the speed is great enough for the force of air resistance to balance the

weight of the sky diver; the net force and (consequently) the acceleration drop to zero. This

constant descent velocity is known as the terminal velocity. For a sky diver falling in the

spread-eagle position without a parachute, the value of the proportionality constant kin the

drag equationFdrag = kv2 is approximately kg/m. Therefore, if the sky diver has a total

mass of 70kg (which corresponds to a weight of about 150 pounds), her terminal velocity is

Or approximately 120 miles per hour.

Once the parachute opens, the air resistance force becomes Fair resist =Kv, and the equation of

motion (*) becomes

or more simply,

whereB =K/m. Once the parachutist's descent speed slows to v =g/B = mg/K, the preceding

equation says dv/dt= 0; that is, v stays constant. This occurs when the speed is low enough

for the weight of the sky diver to balance the force of air resistance; the net force and

(consequently) the acceleration reach zero. Again, this constant descent velocity is known as


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the terminal velocity. For a sky diver falling with a parachute, the value of the

proportionality constantKin the equationFa ir res is t =Kv is approximately 110 kg/s.

Therefore, if the sky diver has a total mass of 70 kg, the terminal velocity (with the

parachute open) is only

which is about 14 miles per hour? Since it is safer to hit the ground while falling at a rate of

14 miles per hour rather than at 120 miles per hour, sky divers use parachutes.

Example 5: After a free-falling sky diver of mass m reaches a constant velocity ofv1, her

parachute opens, and the resulting air resistance force has strengthKv. Derive an equation

for the speed of the sky divertseconds after the parachute opens.

Once the parachute opens, the equation of motion is

WhereB =K/m. The parameter that will arise from the solution of this first-order

differential equation will be determined by the initial condition v (0) = v1 (since the sky

diver's velocity is v1 at the moment the parachute opens, and the clock is reset to t= 0 at

this instant). This separable equation is solved as follows:


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Now, since v(0) = v1gBv1 = c, the desired equation for the sky diver's speed tseconds

after the parachute opens is

Note that as time passes (that is, as tincreases), the term e( K/m)tgoes to zero, so (as

expected) the parachutist's speed v slows to mg/K, which is the terminal speed with the

parachute open.

Other application :-

(A) ELECTRICAL CIRCUIT:

The equation of electric circuit depends upon the

following laws

1) I = dq/dt

2) Voltage drop across resistance R = Ri

3) Voltage drop across inductance L = Ldi/dt

4) Voltage drop across capacitance C = q/C

(A) MECHANICS : it can be used to drive an equation

of mechanics using differential equation.


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(B) HEAT CONDUCTION : IT is used to find the

temperature of body and many other places where

temperature level needs to be known.

(C) CHEMICAL ACTIONS : TO find the amount mass

converted from one form to another. Effect of

temperature on water.

NEWTONS LAW OF COOLING - the rate of decrease of

the temperature is proportional to the difference between

the temperature of the body and that of the medium.

(E) SPRINGS IT can be used to find out tension of the

string or the velocity at which the spring is moving in to

and fro motion and its acceleration

REFERENCES:

http://www.cliffsnotes.com/study_guide/Applications-of-FirstOrder-Equations.topicArticleId-

19736,articleId-19734.html#ixzz0mksKvsQC


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Thomas, G. B., Jr. and Finney, R. L. Calculus and Analytic Geometry, 8th ed. Reading, MA:

Addison-Wesley, 1996.

http//Exact differential - Wikipedia, the free encyclopedia.mht

http://Exact Differential -- from Wolfram MathWorld.mht
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